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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 2 Subquestion (i) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 2 Subquestion (i) Maths Textbook Solution.

Answers (1)

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Answer:
\frac{-\pi }{3}
Hint:
2\sin^{-1}x= \sin^{-1}\left ( 2x\sqrt{1-x^{2}} \right )
Given:
Find\, principal \, value\, o\! f \sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}}
Solution:
\sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}}
= \sin^{-1}\frac{1}{2}-\sin^{-1}\left ( 2\times \frac{1}{\sqrt{2}}\sqrt{1-\left ( \frac{1}{\sqrt{2}} \right )^{2}} \right )              
= \sin^{-1}\frac{1}{2}-\sin^{-1}1
= \frac{\pi }{6}-\frac{\pi }{2}
= \frac{-\pi }{3}
There\! fore, principal\, value\, o\! f \sin^{-1}\frac{1}{2}-2\sin^{-1}\frac{1}{\sqrt{2}} \, is = \frac{-\pi }{3}.

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